Proposition 6

Let X and Y be topological spaces and let $f: X\rightarrow Y$ be a continuous and relatively connected mapping. Suppose that every point of Y possesses an open neighbourhood V whose inverse image $\overset{-1}{f}(V)$ is simply connected. Then, for every covering Z of X, there exists a covering T of Y such that $X\times_YT$ is X-isomorphic to Z.

Let $\mathscr{U}$ be the set of open sets of Y whose inverse image in X

is simply connected. For every $V\in \mathscr{U}$, the covering $\overset{-1}{f}(V)\times_XZ$

of $\overset{-1}{f}(V)$ is trivializable; there thus exists a discrete space $F_V$ and an

isomorphism of coverings $g_V:\overset{-1}{f}(V)\times_XZ\rightarrow \overset{-1}{f}(V)\times F_V$. For every

pair $(V,V')$ of open sets belonging to $\mathscr{U}$, the mapping $f_{V\cap V'}:\overset{-1}{f}(V\cap$ $V')\rightarrow V\cap V'$ is relatively connected. By proposition 5 of IV, p. 356, there exists a unique isomorphism of coverings of $V\cap V'$, $h_{V',V}: (V\cap V')\times F_V\rightarrow (V\cap V')\times F_{V'}$, such that one has $f^*(h_{V',V})(x, t) =$ $g_{V'}(g_V^{-1}(x, t))$ for every $x\in V\cap V'$ and every $t\in F_V$. If V, $V',V''$ are elements of $\mathscr{U}$, one has $h_{V'',V}(x, t) =h_{V'',V'}(h_{V',V}(x, t))$ for every $x\in V\cap V'\cap V''$ and every $t\in F_V$. There then exists a unique Y-space T and, for every $V\in \mathscr{U}$, an isomorphism $h_V: T_V\rightarrow V\times F_V$, such that one has $h_{V',V}(x, t) =h_{V'}\circ h^{-1}_V(x, t)$ for every pair $(V,V')$ of open sets belonging to $\mathscr{U}$, every $x\in V\cap V'$ and every $t\in F_V($cf. TG, I, p. 16). The space T is in particular a covering of Y. There moreover exists a

unique mapping of $X\times_YT$ onto Z whose restriction to $\overset{-1}{f}(V)\times_YT$ is given by $g^{-1}_V\circ f^*(h_V)$ and this is an isomorphism of X-spaces, whence the proposition.